Integrating ln(x) for Finding Area

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Homework Help Overview

The discussion revolves around the integration of the natural logarithm function, ln(x), specifically in the context of finding the area under the curve. Participants are exploring methods of integration within the subject area of Calculus.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of Integration by Parts as a method for integrating ln(x), with some expressing uncertainty about their familiarity with this technique. Others suggest alternative approaches, such as using Simpson's Rule or integrating with respect to y instead of x.

Discussion Status

The discussion is active, with various methods being proposed and explored. Some participants express confusion or lack of knowledge regarding certain integration techniques, while others provide guidance on different approaches. There is no clear consensus on the best method to use.

Contextual Notes

Some participants note that the original poster is in a school setting and may have constraints regarding their current understanding of calculus concepts, which influences their approach to the problem.

ekinnike
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i can't find it anywhere. please help. i need to kno the inegration of lnx to find an area. please help thx
 
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Integration is a Calculus topic, and therefore, should not be posted in the Precalculus section. :smile:
Anyway, to integrate ln(x), we use Integration by Parts (have you covered Integration by Parts yet?), i.e:
[tex]\int u dv = uv - \int v du[/tex]
We often use Integration by Parts, when no other methods can solve the integral.
So, we want to integrate this:
[tex]\int \ln (x) dx[/tex]
We then let u = ln(x), and dv = dx
So that implies du = dx / x, and v = x.
Substitute all those into the formula, we have:
[tex]\int \ln (x) dx = x \ln (x) - \int x \times \frac{dx}{x} = ...[/tex]
Can you go from here? :)
 
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VietDao29 said:
Integration is a Calculus topic, and therefore, should not be posted in the Precalculus section. :smile:
Anyway, to integrate ln(x), we use Integration by Parts (have you covered Integration by Parts yet?), i.e:
[tex]\int u dv = uv - \int v du[/tex]
We often use Integration by Parts, when no other methods can solve the integral.
So, we want to integrate this:
[tex]\int \ln (x) dx[/tex]
We then let u = ln(x), and dv = dx
So that implies du = dx / x, and v = x.
Substitute all those into the formula, we have:
[tex]\int \ln (x) dx = x \ln (x) - \int x \times \frac{dx}{x} = ...[/tex]
Can you go from here? :)
lol i didnt learn those yet. BUt I am glad it more complicated than it look. NO i can't do it. BUt i really appriciated the help. Instead i will try use simpsons rule. the question is area under the curve. 2day is monday-im at school school today so i quickly ask one of the mathematics teacher. =) i think he also hesitated say u can solve them by harder way but instead he told me simpson rule
 
Last edited:
It can be done without integration by parts. Try integrating over y instead of x, that is consider your graph to be x=exp(y).
 
You can't integrate ln(x) by parts!
 
u=lnx dv= dx
du= 1/x dx v=x

=xlnx-integral(1)dx
= xlnx-x+c

lnx using IBP:)
 
This site is great for checking your work when integrating:

http://integrals.wolfram.com

It will not give you fully worked out answers but it will give you the final answer. Of course, your teacher will want you to solve problems showing your work all the way through, but this is still a great site.
 

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